Prove that $g \circ f$ is a one-to-one function

Question

Let $f$ and $g$ be one-to-one functions such that the domain of $f$ is $A$, the range of $f$ is $B$, the domain of of $g$ is $B$, and the range of $g$ is $C$.

1. Prove that $g \circ f$ is a one-to-one function.
2. Prove that $f^{−1} \circ g^{−1}$ is the inverse of $g \circ f$.

Not sure if I should be letting $f(x)$ and $g(x)$ be actual functions here then proving it, or what's going on..

Answer 1

Here's a simple proof for part 1:

Suppose $h(x)=g(f(x))$

Since we know $f,g$ are two (equal or distinct) one-to-one functions, for all $x\ne y \in A$, $f(x) \neq f(y)$ (by definition of one-to-one).

Now, we are given that $f(x), f(y) \in B$, which is also the domain of $g$.

Also, we now know that $g(f(x)) \neq g(f(y))$, since we know that $g$ is also a one-to-one function.

But we have $h(x)=g(f(x)), h(y) =g(f(y)) \implies h(x) \neq h(y)$, which is the definition of a one-to-one function.

This proves our theorem, since $h(x) =g(f(x)) = g\circ f$.

Answer 2

The goal: We can prove $g∘f$ is one-to-one by using the hypothesis to shows that $g(f(x)) = g(f(y)) \implies f(x) = f(y)$

The hypotheses are:
$ f: A \to B$ and $f$ is injective. (injective mean one-to-one)
$ g: B \to C$ and $g$ is injective.

Proof: Let $f$ and $g$ be an injective function. Assume there exist two elements x and y in A such that $x \neq y$, then $f(x) \neq f(y)$ because $f$ is injective. So $f(x)$ and $f(y)$ are two distinct elements in B. Thus, $g(f(x)) \neq g(f(y))$ are two distinct elements in $C$ because g is injective. We have shown that: $$f(x) \neq f(y) \implies g(f(x)) \neq g(f(y))$$ which equivalent to: $$g(f(x)) = g(f(y)) \implies f(x) = f(y)$$ $\therefore g\circ{f}$ is a one-to-one function.

$$\tag*{$\blacksquare$}$$

If you struggle, comment the part that you don't understand about this proof.